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Make "S" the subject of formula in V = sqrt(U2 + 2AS)

Mathematics

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Answer

C

Step-by-step solution

23 steps

Part 1· 2 steps

  1. Break down the number 999,999,999,995 into place value groups.

    The number is 999 billion, 999 million, 999 thousand, and 995.

  2. Write the number in words based on the place value groups.

    Nine hundred and ninety-nine billion, nine hundred and ninety-nine million, nine hundred and ninety-nine thousand and ninety-five.
    The correct option is C.

    • C. Nine hundred and ninety-nine billion, nine hundred and ninety-nine million, nine hundred and ninety-nine thousand and ninety-five

Part 2· 3 steps

  1. Find the H.C.F (Highest Common Factor) for the first set of numbers

    $9, 12, 15, 24$.
    Factors of 9: $1, 3, 9$
    Factors of 12: $1, 2, 3, 4, 6, 12$
    Factors of 15: $1, 3, 5, 15$
    Factors of 24: $1, 2, 3, 4, 6, 8, 12, 24$
    The common factors are $1, 3$. The H.C.F is 3.

  2. Find the H.C.F for the second set of numbers

    $9, 12, 15$.
    Factors of 9: $1, 3, 9$
    Factors of 12: $1, 2, 3, 4, 6, 12$
    Factors of 15: $1, 3, 5, 15$
    The common factors are $1, 3$. The H.C.F is 3.

  3. Find the product of the two H.C.F values.

    Product =3×3=9= 3 \times 3 = 9.
    The correct option is C.

    • C. 9

Part 3· 3 steps

  1. Identify the total volume of palm oil and the volume each bottle holds.

    Total volume =210= 210 litres.
    Volume per bottle =34= \frac{3}{4} litres.

  2. Divide the total volume by the volume per bottle to find the number of bottles.

    Number of bottles=Total volumeVolume per bottle=21034\text{Number of bottles} = \frac{\text{Total volume}}{\text{Volume per bottle}} = \frac{210}{\frac{3}{4}}

  3. Perform the division.

    Number of bottles=210×43=210×43=70×4=280\text{Number of bottles} = 210 \times \frac{4}{3} = \frac{210 \times 4}{3} = 70 \times 4 = 280

    The correct option is B.

    • B. 280

Part 4· 3 steps

  1. Find the H.C.F (Highest Common Factor) of $9, 12,$ and 18.

    Prime factorization:
    9=329 = 3^2
    12=22×312 = 2^2 \times 3
    18=2×3218 = 2 \times 3^2
    The common prime factor is 3, and the lowest power is 313^1. So, H.C.F =3= 3.

  2. Find the L.C.M (Lowest Common Multiple) of $9, 12,$ and 18.

    The prime factors involved are 2 and 3.
    Highest power of 2 is 222^2.
    Highest power of 3 is 323^2.
    L.C.M =22×32=4×9=36= 2^2 \times 3^2 = 4 \times 9 = 36.

  3. Find the product of the L.C.M and H.C.F.

    Product =36×3=108= 36 \times 3 = 108.
    The correct option is A.

    • A. 108

Part 5· 2 steps

  1. Recognize that the angles 60∘60^\circ, 4x4x, and 2x2x lie on a straight line.

    The sum of angles on a straight line is 180∘180^\circ.

  2. Set up the equation and solve for xx.

    60∘+4x+2x=180∘60^\circ + 4x + 2x = 180^\circ 60∘+6x=180∘60^\circ + 6x = 180^\circ 6x=180∘−60∘6x = 180^\circ - 60^\circ 6x=120∘6x = 120^\circ x=120∘6x = \frac{120^\circ}{6} x=20∘x = 20^\circ

    The correct option is C.

    • C. 20∘20^\circ

Part 6· 3 steps

  1. Visualize the problem as a right-angled triangle.

    The ladder is the hypotenuse (66m). The angle with the wall is 30∘30^\circ. We need to find the distance of the foot of the ladder from the wall, which is the side opposite to the 30∘30^\circ angle.

  2. Use the sine trigonometric ratio.

    sin⁡(θ)=oppositehypotenuse\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} sin⁡(30∘)=distance6\sin(30^\circ) = \frac{\text{distance}}{6}

  3. Solve for the distance.

    distance=6×sin⁡(30∘)\text{distance} = 6 \times \sin(30^\circ)

    Since sin⁡(30∘)=0.5\sin(30^\circ) = 0.5:

    distance=6×0.5=3.00 m\text{distance} = 6 \times 0.5 = 3.00 \text{ m}

    The correct option is C.

    • C. 3.00m

Part 7· 3 steps

  1. Identify the given dimensions and properties of the trapezoid.

    Top side (BC) =8= 8 cm.
    Bottom side (AD) =16= 16 cm.
    Height =3= 3 cm.
    Sides AB and CD are equal, indicating an isosceles trapezoid.

  2. Find the length of the non-parallel sides (AB and CD).

    Draw perpendiculars from B and C to AD, let them meet AD at E and F respectively.
    Then BCFE is a rectangle, so EF =8= 8 cm.
    Since it's an isosceles trapezoid, AE =FD=AD−EF2=16−82=82=4= \text{FD} = \frac{\text{AD} - \text{EF}}{2} = \frac{16 - 8}{2} = \frac{8}{2} = 4 cm.
    Consider the right-angled triangle ABE. The height BE =3= 3 cm, AE =4= 4 cm.
    Using the Pythagorean theorem:

    AB2=AE2+BE2AB^2 = AE^2 + BE^2 AB2=42+32AB^2 = 4^2 + 3^2 AB2=16+9AB^2 = 16 + 9 AB2=25AB^2 = 25 AB=25=5 cmAB = \sqrt{25} = 5 \text{ cm}

    So, CD =5= 5 cm.

  3. Calculate the perimeter.

    Perimeter =AB+BC+CD+AD= \text{AB} + \text{BC} + \text{CD} + \text{AD}
    Perimeter =5+8+5+16=34 cm= 5 + 8 + 5 + 16 = 34 \text{ cm}
    The correct option is D.

    • D. 34cm

Part 8· 4 steps

  1. Start with the given formula.

    V=U2+2aSV = \sqrt{U^2 + 2aS}

  2. Square both sides to eliminate the square root.

    V2=U2+2aSV^2 = U^2 + 2aS

  3. Isolate the term containing S by subtracting U2U^2 from both sides.

    V2−U2=2aSV^2 - U^2 = 2aS

  4. Make S the subject by dividing both sides by $2a$.

    S=V2−U22aS = \frac{V^2 - U^2}{2a}

    The correct option is C.

    • C. S=V2−U22aS = \frac{V^2 - U^2}{2a}
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